Is this an acceptable definition for $\sqrt{a}$, where $a \in \mathbb{R}$?

If $a\geq 0, \sqrt{a} = b \in \mathbb{R}$ s.t. $b\geq 0, b^2 = a$.

I'm proving some theorems involving $\sqrt{a}$, in the context of introductory real analysis.

  • $\begingroup$ How have you defined $b^2$? $\endgroup$ – Don Larynx Nov 1 '13 at 5:16
  • $\begingroup$ This definition looks fine, except on the first line you write "Is this an acceptable definition for $\sqrt{a}$, where $a\in\mathbb{R}$", you probably $a\in\mathbb{R}$ want to be $\{x\in\mathbb{R}:x\geq 0\}$. The other point is stylistic. I think that the term "such that" in the second line would work better as an if and only if. I do not recall seeing such that used in that way (although it might be done for all I know). $\endgroup$ – Baby Dragon Nov 1 '13 at 5:55
  • $\begingroup$ Yes, that definition can be decoded, but I think it would be better to use more English words. For example, you could say "given a real number $a \ge 0$ we define $\sqrt{a}$ to be the unique real number $b \ge 0$ satisfying $b^2=a$." $\endgroup$ – Trevor Wilson Nov 1 '13 at 6:01
  • $\begingroup$ @TrevorWilson How have you defined $b^2$? $\endgroup$ – Don Larynx Nov 1 '13 at 6:53
  • $\begingroup$ @DonLarynx I haven't defined it above, but when I define $b^2$ I define it as $b \times b$ (or $b \cdot b$, depending on how multiplication is written in the context.) Or I might define $b^n$ by recursion on natural numbers $n$ as follows: $b^0 = 1$, and $b^{n+1} = b^n \cdot b$. $\endgroup$ – Trevor Wilson Nov 1 '13 at 15:17

Let $a, b \in \mathbb{R}$. $\sqrt{a}$ is the square root of $a$ and is defined as $f(a) = b$ such that $f:a \to a\times a = b$. Notice the product then can never be negative, because $\mathbb{R}$ is reflexive.

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    $\begingroup$ I'm afraid I don't understand this definition. What is $f$? Shouldn't $\sqrt{a}$ be the unique something rather than "all" of something? And what do you mean by "$\mathbb{R}$ is reflexive"? $\endgroup$ – Trevor Wilson Nov 1 '13 at 6:08
  • $\begingroup$ Reflexive means $a \tilde R a$ $\endgroup$ – Don Larynx Nov 1 '13 at 6:33
  • $\begingroup$ WHY THE DOWNVOTE? $\endgroup$ – Don Larynx Nov 1 '13 at 6:51
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    $\begingroup$ Neither downvote was mine, but presumably they happened because your answer doesn't make any sense. I still don't understand why you are talking about an $f$, and also after your latest edit it seems to say that $\sqrt{a} = a \times a$ rather than the other way around. Also, I know what it means for a binary relation $R$ to be reflexive, but I still don't know that it means for the ordered field $\mathbb{R}$ of real numbers to be reflexive, or what that would have to do with the question. $\endgroup$ – Trevor Wilson Nov 1 '13 at 15:13
  • $\begingroup$ $=$ is the equality relation and implies either direction @TrevorWilson. $\endgroup$ – Don Larynx Nov 1 '13 at 21:27

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